{4y^2+4y+1 <=2y+7</p>
{ 4sin^2y+ (2-2V2)siny-V2>=0
siny=t
4t^2 + ( 2 -2V2)t-V2 >=0
D=(2-2V2)^2+4*4*V2 =4-8V2+8+16V2=8V2+12
t=(-2+2V2+V(8V2+12)/8~0.707
t2=(-2+2V(2)-V(8sqrt(2)+12))/8 =1/2
----------------------------------------------->y
- 0.5 - 0.707 +
[ (-2+2V2+V(8V2+12)/8;+oo)
4y^2+4y+1 <=2y+7 </p>
4y^2+2y-6<=0</p>
D=4+4*4*6 =V100= 10
y=(-2+10)/8=1
y2=(-2-10)/8=-12/8=-3/2
----------------------------------------------->y
-1.5 0,707 1
Ответ [ (-2+2V2+V(8V2+12)/8; 1]